Recent content by pandatime

Thank you both of you! It's actually the first time I've done one of these problems, I usually deal with other scenarios and so I think I got tripped up by the fact that they asked for work done by gravity instead of just like an applied force that a person does in most of the other questions, I...

Hi! So, I've actually already solved this problem.. for the most part. I have split up the work into two sections, floor 0 to 10, and floor 10 to 15. From floor 0 to 10, I did ## F_{elevator} = w_{pass.} + w_{elev.} ## ## F = (70)(20 (num. of pass.))(9.8) + (800)(9.8) ## ## F_{elev.} = 21560N...

Hi guys! sorry I went to bed last night and also just got home so I didn't see these replies... that makes sense, basically translating it into a problem where there are two numerical points we can use to calculate centre of mass from... I'll try that right now. I'm not sure I understand how...

omg sorry it is like 5am here right now I'm tripping HAHA thank you, so I guess this is the answer even if it is slightly off? I just worry because this is just practice for me to understand concepts, I don't really care about the answer because it's not for marks but I just don't want it to be...

unfortunately the answer is 2.32kN so there must be something missing... it's quite close tho... I am glad we are on the same page though i felt quite lost

ok so i did ## \int_0^{4.88} 204 dx + \int_{4.88}^{16.7} 111 dx ## and I got 2307N? it's still like 200N off the answer but am I on the right track?

Yes that was what I was thinking, I expressed it in my original post, so I think there is a miscommunication and I am confused, trying my best to understand I think i was just confused as to what haruspex was saying because it was talking about how the centre of mass don't move, I am still not...

wait so this whole problem is just a trick question and the centre of mass is in the centre of the rod just like in a uniform rod problem? and nothing has changed due to density?

but wouldn't it be different for this problem because half of the rod is denser than the other half of the rod?

I'm not quite sure. Would that force matter though since the problem asks for the tension at 16.7 metres? i'm not sure if 'constant' implies that k is changing throughout the problem even past 4.88 metres? ##k = \frac{d T}{d x}## how do we find ##dT## if we don't know any value of the tension...

would there not be a problem since we don't have mass?

I used it because the problem says that once ##x > 4.88## then the spring constant ##k## is ##111 N/m##

the definition of ##\frac{\Delta T}{\Delta x}##? I'm not sure I follow, wouldn't that just lead to the same formula of ##k = \frac{\Delta T}{\Delta x}## ##111 N/m = \frac{\Delta T}{16.7m}## which was what I did in my attempted answer above which was unfortunately not the answer Correct me if...

I feel like I'm missing something fundamental here. Given only the lengths and the densities, how am I supposed to find a numerical centre of mass?Thought process so far: Are we supposed to use the ratio of the densities to find this answer? like ##\frac{8g/cm^3}{2.7g/cm^3}##? and then use that...

ahhh sorry to ask, but english isn't my first language, what is superpose?